智慧职教: 已知二元复合函数,则. 01cb6b87a8617df331f072ec332b69eb 5f5352d0b8855bb3a7b50d8c1f2cb498 79103e721a0eb9c2f09d59b9a90cfb1e
举一反三
- 已知函数f(x)=ax+a-x(a>0且a≠1),且f(1)=3,则f(0)+f(2)的值是( ) A: 7 B: 8 C: 9 D: 10
- 智慧职教: 已知二元复合函数,则. 01cb6b87a8617df331f072ec332b69eb 5f5352d0b8855bb3a7b50d8c1f2cb498 79103e721a0eb9c2f09d59b9a90cfb1e
- 智慧职教: 已知二元复合函数,则. 01cb6b87a8617df331f072ec332b69eb 5f5352d0b8855bb3a7b50d8c1f2cb498 79103e721a0eb9c2f09d59b9a90cfb1e
- 智慧职教: 已知二元复合函数,则. 01cb6b87a8617df331f072ec332b69eb 5f5352d0b8855bb3a7b50d8c1f2cb498 79103e721a0eb9c2f09d59b9a90cfb1e
- 下图所示机构自由度计算,( )是正确的。 A: mg src="http://p.ananas.chaoxing.com/star3/origin/cb07ca0fb12be985c301490389c1e187.jpg" B: F=3×7 –(2×9 + 2 – 2)– 2 = 1 C: F=3×7 –(2×9+ 2– 0)– 0 = 1 D: F=3×7 –(2×8+ 2 – 0)– 2 = 1 E: F=3×5 –(2×6+ 2– 0)– 0 = 1